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[英]MySQL get all rows except first, for all rows that share the same field values
[英]how to get all rows except with specific values
您好,有沒有辦法只獲取組織尚未使用的survey_name
?
`survey table`
+-----------+-------------+
| survey_id | survey_name |
+-----------+-------------+
| 1 | name1 |
| 2 | name2 |
| 3 | name3 |
| 4 | name4 |
| 5 | name5 |
+-----------+-------------+
`link table`
+---------+-------------+-----------------+
| link_id | survey_link | organisation_id |
+---------+-------------+-----------------+
| 1 | 1(survey_id)| 1 |
| 2 | 2 | 1 |
| 3 | 2 | 2 |
| 3 | 3 | 2 |
| 3 | 6 | 2 |
+---------+-------------+-----------------+
在此數據庫結構中,您會看到每個組織的信息:
可用的調查組織1:
可用的調查組織1:
我試過使用WHERE NOT (survey_id= $row['survey_id'])
//get all survey's that are in use
$sqlCheckLink = "SELECT `survey_id` FROM `link_info` WHERE `organisation_id`=".$_SESSION['organisation_id']."";
$resultCheck = mysqli_query($conn, $sqlCheckLink);
if ($resultCheck ->num_rows > 0) {
while ($id = $resultCheck -> fetch_assoc()) {
//show all surveys that are available
$sqlGetSurveys = "SELECT * FROM `survey_info` WHERE NOT (survey_id = ".$id['survey_id'].")";
$resultAllSurveys = mysqli_query($conn, $sqlGetSurveys);
if ($resultAllSurveys ->num_rows > 0) {
while ($row = $resultAllSurveys-> fetch_assoc()) {
//echo content
}
}
}
}
從這里但它似乎不起作用......使用這種方法我也得到了正在使用的調查。
如果有人可以提供幫助,將不勝感激!
只需使用左連接:
SELECT t.survey_name
FROM survey_table as t
LEFT JOIN link_table AS l ON l.survey_link = t.survey_id
WHERE l.link_id IS NULL
未經測試,但您可以了解這個想法。 我希望這對你有幫助。
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